Mathematics, 18.01.2020 05:31, bvghchg4401

An examiner sets twelve problems, and tells the class that the exam will consist of six of them, selected at random. gavin memorises the solutions to eight problems in the list, but cannot solve any of the others. what is the chance he gets four or more correct?

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answered: MadiAbbott798

The chances Gavis get four or more correct problems is 8/11 or 72.72%

Step-by-step explanation:

The exam is composed of 6 problems out of 12 possible cases (Pc=12). There are 2 groups of problems:

The 8 problems that Gavin has the answer (Problems A).

The 4 problems that Gavin hasn´t the answer (problems B).

Therefore:

P(A≥4)= P(A=4) ∪ P(A=5) ∪ P(A=6) = P(A=4) + P(A=5) + P(A=6)

Before we start analyzing the problem, we have to understand that problems in the exam are selected at random, but a problem can´t be selected twice. therefore picking a specific problem will reduce the pool of that specific group and of the total number of available problems.

If we call $X_{ij}$ to the probability of an answer of the X group to be the i° picked problem from the j° picked problem of that group:

$X_{ij}=\frac{N_{GRX}-j+1}{12-i+1}$ with $N_{GRX}$ the total number of problems in that group.

We analyze now 3 different problems:

$P(A=6)=A_{11} \cdot A_{22} \cdot A_{33} \cdot A_{44} \cdot A_{55} \cdot A_{66}\\P(A=6)=\frac{8}{12} \frac{7}{11} \frac{6}{10} \frac{5}{9} \frac{4}{8} \frac{3}{7} = \frac{1}{33}$

$P(A=5)=A_{11} \cdot A_{22} \cdot A_{33} \cdot A_{44} \cdot A_{55} \cdot B_{61}+ A_{11} \cdot A_{22} \cdot A_{33} \cdot A_{44} \cdot B_{51} \cdot A_{65}+...+B_{11} \cdot A_{21} \cdot A_{32} \cdot A_{43} \cdot A_{54} \cdot A_{65}\\$

$P(A=5)=\frac{8}{12} \frac{7}{11} \frac{6}{10} \frac{5}{9} \frac{4}{8} \frac{4}{7} + \frac{8}{12} \frac{7}{11} \frac{6}{10} \frac{5}{9} \frac{4}{8} \frac{4}{7} + ... + \frac{4}{12} \frac{8}{11} \frac{7}{10} \frac{6}{9} \frac{5}{8} \frac{4}{7} = 6 \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot4 \cdot 4}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}=\frac{8}{33}$

For P(A=4) we can take the solution from P(A=5) and say that:

$P(A=4)=c \cdot \frac{4 \cdot 3 \cdot 8 \cdot 7 \cdot 6 \cdot5 }{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}$ where "c" is the combinatorial of 2 problems B with 4 problems A. In this case "c" is 15, therefore:

$P(A=4)=\frac{8}{11}$

$P(A \ge4)= P(A=4) + P(A=5) + P(A=6) =\frac{1}{33} +\frac{8}{33} +\frac{5}{11} = \frac{8}{11}$

answered: Guest

step-by-step explanation:

answered: Guest

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.